Question: Simplify the following expression: $y = \dfrac{-7x^2+24x- 9}{-7x + 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(-9)} &=& 63 \\ {a} + {b} &=& &=& {24} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $63$ and add them together. The factors that add up to ${24}$ will be your ${a}$ and ${b}$ When ${a}$ is ${3}$ and ${b}$ is ${21}$ $ \begin{eqnarray} {ab} &=& ({3})({21}) &=& 63 \\ {a} + {b} &=& {3} + {21} &=& 24 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 +{3}x) + ({21}x {-9}) $ Factor out the common factors: $ x(-7x + 3) - 3(-7x + 3)$ Now factor out $(-7x + 3)$ $ (-7x + 3)(x - 3)$ The original expression can therefore be written: $ \dfrac{(-7x + 3)(x - 3)}{-7x + 3}$ We are dividing by $-7x + 3$ , so $-7x + 3 \neq 0$ Therefore, $x \neq \frac{3}{7}$ This leaves us with $x - 3; x \neq \frac{3}{7}$.